I am reading chapter 3 of Folland's Real Analysis Book, where it introduces the Radon-Nikodym Derivative of a sigma finite signed measure with respect to another sigma finite positive measure.
Let $\nu$ be a $\sigma$-finite signed measure and $\mu$ be a $\sigma$-finite positive measure on $(X,M)$. There exists unique $\sigma$-finite signed measure $\lambda,\rho$ on the space such that $$\lambda \perp \mu \text{ } \rho << \mu \text{ and } \nu=\lambda +\rho$$ and there is an $\mu$-integrable function $f:X\to \mathbb{R}$ such that $d\rho=f d\mu$ where $f$ is the Radon-Nikodym derivative.
I am just wondering if this is somehow related to the derivative in calculus in anyway, since Riemann integral of a continuous fct on bounded interval is equal to Lebesgue integral and Lebesgue measure is $\sigma$-finite?
Or maybe this question is kind of stupid and they are not related in any way.
Best Answer
The two are very much related.
You may be familiar with Lebesgue's Differentation Theorem. In the special case where $\lambda$ is the Lebesgue measure in $\mathbb{R}$, and for a $\sigma$-finite measure $\nu$ with $\nu << \lambda$ we have $$\frac{d\nu}{d\lambda}(x)=\lim\limits_{h\to 0}\frac{\int\limits_{x-h}^{x+h}{\frac{d\nu}{d\lambda}(y)}d\lambda(y)}{2h}= \lim\limits_{h\to 0}\frac{\nu([x-h,x+h])}{2h} \quad \text{$\lambda$-a.e.}$$
And this in turn is the differential quotient of the function $f(x)=\nu((-\infty,x])$
A more general version of the theorem holds for Radon measures. It can be shown that if $\nu << \mu$, then $$\frac{d\nu}{d\mu}(x)=\lim\limits_{r \to 0} \frac{\nu({B_{r}(x)})}{\mu(B_{r}(x))} \quad \text{$\mu$-a.e.}$$
see Theorem 1.30 in
Evans, Lawrence Craig; Gariepy, Ronald F., Measure theory and fine properties of functions, Textbooks in Mathematics. Boca Raton, FL: CRC Press (ISBN 978-1-4822-4238-6/hbk). p.50 (2015). ZBL1310.28001.